The question might at first sight sound like the answer is trivially "yes", so let me clarify the question a bit. Consider given a nonnegative Radon measure $\mu$ on $\mathbb{R}^n$. Let $\mathcal{D}(\mathbb{R}^n)$ denote the space of real test functions, that is, the space of infinitely differentiable functions with compact support from $\mathbb{R}^n$ to $\mathbb{R}$. We may then define the operator $T:\mathcal{D}(\mathbb{R}^n)\to\mathbb{R}$ by $$ T(\varphi) = \int_{\mathbb{R}^n} \varphi(x) d\mu(x). $$ This is then a distribution, in the sense of being a continuous linear functional $\mathcal{D}(\mathbb{R}^n)\to\mathbb{R}$, when $\mathcal{D}(\mathbb{R}^n)$ is given its usual topology based on a family of norms as in Rudin's "Functional analysis".
Now let $\mathcal{S}(\mathbb{R}^n)$ denote the space of real Schwartz functions, meaning the space of rapidly decreasing functions, endowed with its usual topology, again see Rudin's book for details.
$\textbf{Assume the following:}$ That $T$ can be extended to a linear functional on $\mathcal{S}(\mathbb{R}^n)$ which is continuous in the topology of $\mathcal{S}(\mathbb{R}^n)$.
$\textbf{My question is this:}$ Does it holds that $$ T(\varphi) = \int_{\mathbb{R}^n} \varphi(x) d\mu(x). $$ for all $\varphi\in\mathcal{S}(\mathbb{R}^n)$?
$\textbf{Some remarks:}$ The problem is that while we know that $T$ extends from $\mathcal{D}(\mathbb{R}^n)$ to $\mathcal{S}(\mathbb{R}^n)$, we do not know that the integral form of the operator carries over from $\mathcal{D}(\mathbb{R}^n)$ to $\mathcal{S}(\mathbb{R}^n)$. In fact, we do not even know that all functions in $\mathcal{S}(\mathbb{R}^n)$ are integrable with respect to $\mu$.
A natural first approach to the problem would be to consider some kind of approximation argument. For example, take $\varphi\in\mathcal{S}(\mathbb{R}^n)$. Assume that $\varphi\ge0$. Using convolutions, we may then construct a sequence $(\varphi_n)$ in $\mathcal{D}(\mathbb{R}^n)$ which converges monotonely and in $\mathcal{S}(\mathbb{R}^n)$ to $\varphi$. This then yields
$$ \int_{\mathbb{R}^n} \varphi(x) d\mu(x) = \lim_n \int_{\mathbb{R}^n} \varphi_n(x) d\mu(x) = \lim_n T(\varphi_n), $$ where the limit a priori may be infinite. However, as we have assumed that $T$ extends to $\mathcal{S}(\mathbb{R}^n)$, the limit of $T(\varphi_n)$ is finite, and so $\varphi$ is integrable with respect to $\mu$. This should show that the integral form of the operator is preserved for all nonnegative Schwartz functions.
For a general Schwartz function $\varphi$, the natural thing would be to write $\varphi = \varphi^+-\varphi^-$, where $\varphi^+$ and $\varphi^-$ are the positive and negative parts of $\varphi$, respectively. However, while these two functions are nonnegative, they are not Schwartz functions, nor test functions for that matter, and so we cannot apply $T$ to them.